2.3 Exercises from the Feyman Lectures on Physics

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new90
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Homework Statement
a body is acted upon by n forces and is in static equilibrium.Use the principle of virtual work t prove that
if n = 1 the magnitude of the force must be zero
Relevant Equations
no equation
i don't know what's the answer but I think that the force needs to be gravity because is going down to the floor
 
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new90 said:
i don't know what's the answer but I think that the force needs to be gravity because is going down to the floor
What if the body is the middle of outer space?
 
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then there willl be work because there there will be a force pusing in some direction
 
Please try to understand what Professor Feynman meant by the words 'static equilibrium'.
 
i think it means that the body its not moving
 
new90 said:
i think it means that the body its not moving
It means that any force acting on the body in any direction is counter-balanced by a force acting in an oppositional direction.
 
Also remember that the principle of virtual work says the virtual work must be zero for any infinitesimal displacement. Because you could easily choose ##d\vec{r}## to be orthogonal to the resultant force!
 
etotheipi said:
Also remember that the principle of virtual work says the virtual work must be zero for any infinitesimal displacement. Because you could easily choose ##d\vec{r}## to be orthogonal to the resultant force!
Well, equally true is that it must be non-zero for any non-zero force -- if it is infinitesimal it is still infinitesimally non-zero. Let's please not abuse zero while we are petting our pet inconsistencies . . .
 
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Are you able to do it just by condition implied by static equilibrium? I mean do you know that ##\Sigma## notation for statics?
 
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sysprog said:
Well, equally true is that it must be non-zero for any non-zero force -- if it is infinitesimal it is still infinitesimally non-zero. Let's please not abuse zero while we are petting our pet inconsistencies . . .

My apologies, I was being careless. I meant to say "for any virtual displacement".
 
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Adesh said:
Are you able to do it just by condition implied by static equilibrium? I mean do you know that ##\Sigma## notation for statics?
In my view, the sigma notation, in this instance, refers to sum, I think that I don't quite fully understand your question . . .
 
It means that any force acting on the body in any direction is counter-balanced by a force acting in an oppositional direction.
but there's its says only one force
 
sysprog said:
In my view, the sigma notation, in this instance, refers to sum, I think that I don't quite fully understand your question . . .
Does OP know this ## \sum_{i=1}^{n} \vec{F_i} =0## ?
 
new90 said:
whos OP
OP : Original Poster.
 
new90 said:
I DINT KNOW THANKS
Do you know for Static Equilibrium we should have $$ \sum_{i =1}^{n} \vec{F_i} =0 $$ ?
 
i don't know what it means the n
 
new90 said:
i don't know what it means the n
$$\sum_{i=1}^{n} \vec{F_i} = \vec{F_1}+ \vec{F_2} + \vec{F_3} +... + \vec{F_n} =0$$ Are things a little clearer now?
 
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yes thanks
 
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but in this case its just
Fi= 0
 
new90 said:
but in this case its just
Fi= 0
If ##n=1## we have $$ \sum_{i=1}^{1} \vec{F_i} = F_1 = 0$$ How about now?
 
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new90 said:
yes its clear
thanks
But you don’t have to do it like this 😅
You have to do it by the Principle of virtual work.
 
Principle of virtual work = the force x the virtual
displacement?
 
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thanks I answered with principle of virtual work
 
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Adesh said:
$$\sum_{i=1}^{n} \vec{F_i} = \vec{F_1}+ \vec{F_2} + \vec{F_3} +\dots+ \vec{F_n} =0$$
I 'corrected' your ##\LaTeX## ##-## you had 4 dots instead of 3 ##-## :wink: your math was just fine ##\dots##
 
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sysprog said:
I 'corrected' your ##\LaTeX## ##-## you had 4 dots instead of 3 ##-## :wink: your math was just fine ##\dots##
It seems that you have adjusted it’s height too!
 
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